We're asked to find the angle theta
over the closed interval from zero to 180 degrees
between the following vectors
where vector u is equal
to negative five times the unit vector i
and vector v is equal to the eight times the unit vector i
minus nine times the unit vector j,
which means vector u in component form has an x-component
of negative five and a y-component of zero,
which is sketched here in blue.
And vector v
has an x-component of eight
and a y-component of negative nine,
which is sketched here in red,
which means we're looking for this angle here,
the angle between the two vectors in the closed interval
from zero to 180 degrees.
Notice how there is a larger angle here on the other side.
We'll be finding this angle here
using the formula cosine theta is equal
to vector u dotted with vector v
divided by the product of their magnitudes.
And we can solve this equation for angle theta
by taking the inverse cosine or arccosine
of both sides of the equation,
which gives us this equation here.
And we know the output of inverse cosine is
over the closed interval from zero to 180 degrees
or from zero to pi radians,
which is why we'll be finding the smaller angle here
between the two vectors.
So let's begin by finding the dot product
as well as the magnitude of the two vectors.
So let's first find vector u dot vector v,
which is equal to negative five times eight,
plus zero times negative nine,
which is equal to negative 40.
And now let's find the magnitudes of the two vectors.
The magnitude of vector u is equal to the square root
of the square of negative five plus the square of zero,
which is equal to the square root of 25,
which is equal to five.
And now let's find the magnitude of vector v,
which is equal to the square root
of the quantity eight squared,
plus the square of negative nine,
which is equal to the square root
of 64 plus 81,
which is equal to the square root of 145.
So now we know that cosine theta
is equal to the dot product, which is negative 40,
divided by the product of the magnitudes,
which is five times the square root of 145.
And now solving this equation for theta,
we take the inverse cosine and arccosine
of both sides of the equation,
which gives us theta is equal to inverse cosine
of negative 40,
divided by five square root 145.
And now we'll go to the calculator
and find the angle theta in degrees, as well as radians.
And we are told to round to four decimal places.
So go into the calculator.
Let's first find the angle in degrees.
Let's press the Mode key.
Go down to the fourth row.
Notice how Degree is already highlighted.
If it wasn't, we would highlight it and press Enter.
And now we go back to the home screen by pressing 2nd, Mode.
And now we press 2nd, Cosine for inverse cosine,
and then we enter negative 40 divided by.
Now, we need to be careful here
because the five and the square root are in the denominator,
and therefore we need parentheses around this product.
So we have open parenthesis five,
2nd, x squared for the square root,
145, right arrow,
close parenthesis for the denominator,
and then close parenthesis for the input of inverse cosine,
and Enter.
Notice in the fifth decimal place we have a three,
and therefore we round down
to 131.6335 degrees.
However, to make the entry a little bit easier,
what we can do on this calculator is press 2nd,
Cosine for inverse cosine,
and then if we press Alpha, Y= and select this first option,
notice how it brings up this fraction format.
Using this, we don't have to worry
about putting five square root 145 in parentheses.
We can simply enter negative 40, right arrow,
and then enter five square root 145,
right arrow, right arrow, close parenthesis, and Enter.
And of course we get the same result.
So this format can be a little bit easier to use at times.
And now to find the angle in radians,
we simply have to change the mode.
Press Mode, highlight Radian, press Enter,
go back to the home screen.
We can pull up the previous entry by pressing 2nd, Enter.
And now that we're in radian mode, if we press Enter,
we now have the angle in radians.
And notice here we have a three in the fifth decimal place,
and therefore in radians,
we have approximately 2.2974 radians,
which, again, is this angle here between the two vectors.
I hope you found this helpful.